1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Focal length of thin converging lens

  1. Nov 4, 2015 #1
    1. The problem statement, all variables and given/known data

    An outdoor security camera has a simple lens with focal object length fo = 6.00 cm and perfectly focuses the image of a bird onto its detector located 9.00 cm away from the lens. The animal then flies away from the lens and the new focal distance is 8.00 cm. (a) How far did the bird move? (b) What is the new magnification?

    2. Relevant equations
    $\begin{array}{l}
    \[\frac{1}{f} = \frac{1}{s} + \frac{1}{{s'}}\]
    \end{array}$
    3. The attempt at a solution

    This is my solution:
    $\begin{array}{l}
    \frac{1}{{{f_0}}} = \frac{1}{s} + \frac{1}{{s'}}\\
    \frac{1}{{{s_1}}} = \frac{1}{{{f}}} - \frac{1}{{s'}} = \frac{1}{6} - \frac{1}{9} = \frac{1}{{18}}\\
    {s_1} = 18cm\\
    \frac{1}{{{s_2}}} = \frac{1}{{{f}}} - \frac{1}{{s'}} = \frac{1}{6} - \frac{1}{8} = \frac{1}{{24}}\\
    {s_2} = 24cm\\
    \Delta {s_o} = {s_2} - {s_1} = 24cm - 18cm = 6cm
    \end{array}$
    As my textbook offers limited worked solutions for this topic I have looked around and it seems that the focal length does not change on a 'simple lens', even if it is bi-convex. It is s and s prime that changes. Or am I misunderstanding the initial question? The wording is somewhat vague.

    This is the solution I was presented with:

    f1= 6.00 cm

    s’1 = s’2= 9 cm ( f > s’ >2f ) + image is real, inverted, and reduced (for both)

    f2 = 8.00 cm

    s1 = ?

    s2 = ?

    Ds0= s2- s1

    A camera should mean a bi-convex lens.
    $\begin{array}{l}
    \frac{1}{{{f_0}}} = \frac{1}{s} + \frac{1}{{s'}}\\
    \frac{1}{{{s_1}}} = \frac{1}{{{f_1}}} - \frac{1}{{s'}} = \frac{1}{6} - \frac{1}{9} = \frac{1}{{18}}\\
    {s_1} = 18cm\\
    \frac{1}{{{s_2}}} = \frac{1}{{{f_2}}} - \frac{1}{{s'}} = \frac{1}{8} - \frac{1}{9} = \frac{1}{{72}}\\
    {s_2} = 72cm\\
    \Delta {s_o} = {s_2} - {s_1} = 72cm - 18cm = 54cm
    \end{array}$
    The bird flew 54 cm or just over half a metre from the camera.
     
  2. jcsd
  3. Nov 4, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    For s2, I think the 8 centimeters are supposed to change to 9 cm (so the distance between detector and lens changes). The lens itself does not change.

    Focal length and focal distance are different things.
     
  4. Nov 4, 2015 #3
    Thanks for the reply mfb.

    I think I am also confusing focal length and focal point, but the solution I was given still doesn't make sense to me. If the image remains focused on the detector then the lens must change or the detector-lens distance must change. If the detector-lens distance changes then it would be from 9cm to 8cm because the object distance has definitely changed in the wording of the question.
     
  5. Nov 5, 2015 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Well, you have a good point. A 'simple lens' could not change its focal length.
    However, in this problem it must, in order to keep a sharp image for different object distances s1 and s2.
    So your 'simple' lens has to be a 'zoom' lens. And I think the problem statement is wrong to call it a 'simple lens'.
    https://en.wikipedia.org/wiki/Zoom_lens
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted